AtCoder Beginner Contest 132 F Small Products

Small Products

思路：

整除分块+dp

打表发现，按整除分块后转移方向如下图所示，上面的块的前缀转移到下面的块

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11Z
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<int, pii>
#define puu pair<ULL, ULL>
#define MOD(a, b) (a >= b ? a%b+b : a%b)
#define pdd pair<long double, long double>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 1e5 + 5;
const int MOD = 1e9 + 7;
int n, k, l[N], r[N], cnt = 0;
int dp[105][N];
int main() {
    scanf("%d %d", &n, &k);
    for (int x = 1, y; x <= n; x = y+1) {
        y = n/(n/x);
        l[++cnt] = x;
        r[cnt] = y;
    }
    for (int i = 1; i <= cnt; ++i) dp[1][i] = (dp[1][i-1] + r[i]-l[i]+1)%MOD;
    for (int i = 2; i <= k; ++i) {
        for (int j = 1; j <= cnt; ++j) {
            dp[i][j] = (dp[i][j-1] + (dp[i-1][cnt-j+1])*1LL*(r[j]-l[j]+1)%MOD)%MOD;
        }
    }
    printf("%d
", dp[k][cnt]);
    return 0;
}